Is there a standard to determine Magnification in a print?

<p>It's funny, but I don't know that I have ever seen or heard this addressed in terms of digital photography. As a result, in a rather theoretical project I am working on, I have been a bit stymied trying to figure out the magnification of a final, sized image(print). My results, depending on what factors are used, have been in the millions, hundreds of thousands or tens of thousands of times, all on the same image at the same size.<br /> <br /> With film, we just used the film size. A piece of 35mm film enlarged to 8x12 was considered either 8x linear or 64 times (area) enlargement. With digital is it more complicated? Two sensors of the same size, say for this example FF or 24x36mm, can be 10mp or 20mp for instance. All other things being equal, do you consider an 8x12 print from either the same level of magnification or is there a difference?<br /> <br /> Has anyone ever heard this issue discussed or do you have any thoughts on this?</p>

<p>It's still the original image <em>dimension</em> that sets the amount of magnification. 35mm film is 24x36mm in size. A print made from a 35mm-size sensor would have the same magnification as a 35mm negative would have. Pixel density in the original would certainly affect the number of "pixels per inch" in the resulting print, of course, and that would make it possible to make much larger enlargements.</p>

<p>APS-C film images are 25.1×16.7mm in size, so anything approximating that in sensor size has been called APS-C (approximately 22x15mm, or exactly 22.3x14.9 mm for Canon, and 23.7x15.6mm for Nikon). You simply calculate the increase in linear dimensions of those dimensions just as you would have for 35mm negatives.</p>

<p>JDM, this was actually my last thought in the process although I have been a film photographer for over 30 years--you would think it would have been the first thought! But there seems to be a difference between physical size and they way digital works. As in my example above, assuming nothing changed and the world was perfect---as were lenses, sensors etc--the 10mp capture and the 20mp or maybe a 100mp camera in this perfect world (think no limits in lens resolution, noise, pixel densities and all that technical baggage), would you still consider the magnification from each of these 24x36 sensors the same on that 8x12 print? The physical capture/file isn't the same size to start with--but film was always the same size!?!</p>

<p>I actually came up with the traditional method last and thought it might be the most defensible. But I also considered that if I extract an element and take its ratio, in pixels, to the total file size and then compare that to the print (adjusted for the print dpi) that that may be a more accurate representation.</p>

<p>Just some further thoughts to offer, I guess.</p>

<p>I still am thinking here and would love any more opinions on the matter, but I suppose the sensor size does specify the absolute optical enlargement factor. I find that totally defensible, but still wonder if pixels don't factor in somehow..........</p>

<p>Yes, but think about it. Film differed from print to print. An 8x10 print made from 35mm film exposed and processed for ASA (as we called it) 1000, and a similar print made from the finest grain ASA 32 film were both still 8X linear, regardless of the fact that the grain in the one would have been much coarser than on the other. Pixel or grain density alters the <em>effective</em> size of a potential "high-res" print but it does not alter the linear magnification.<br /> A 1-inch wide image enlarged to 8 inches will still be 8X. I think you're confusing how high quality a print you can get with the issue of linear enlargement.</p>

<p>There are measurements of how high a ppi you can get from a particular image that might be expressed as a ratio of some sort, but those are not measures of <em>size</em> but rather of print <em>quality</em>.</p>

<p>Yes, as I thought about it, I was sort of correlating, in at least a theoretical way, the PPI of an image to how different films resolved an image based on their ISO. So, like I said, optical magnification is probably the most defensible, which gets us right back to a sensor's physical dimensions as the only standard that could be acceptable in such calculations.</p>

<p> </p>

<p>John you can just measure the magnification factor.</p>

<p>Shoot a flat object at a right angle and make two exposures - one with a ruler and one just of your object.<br>

If both images are printed in the same way you can just measure the size of the scale on the ruler in the image by using the same ruler you used for shooting. The ratio of the (ruler image)/(ruler) is the magnification or reduction. The magnification is obviously the same for the image without the ruler. For enhanced precision you may have to control minor parameters as well.</p>

<p>If you shoot 2D objects at an angle you need to apply geometric corrections for the distance and lens aberrations. For 3D objects you may not want to know :-)</p>

<p>In case you really need a "standard" as used in science and technology you need to contact your government's institute of standards. Just for a ratio this is obviously not needed since the same object is used in the image and as object for comparison.</p>

<p>BTW: Starting from "formats" like 24mm x 36mm film or digital sensor is not a good idea. If you read the fine print in your camera manual or measure the exposed area on your film you may notice that the actual size of the exposed area of film or the sensor area may not be exactly this size. And in principle there is no difference between film or digital sensor but the devil is in detail. For example film may not be perfectly flat and the film plane may be different from day to day e.g. in different environment or if the film was transported recently or a long time ago.</p>

<p>The topic is still important today in documentation photography. For general shooting people usually could not care less :-)</p>

<p>Walter, certainly what you describe is particularly important in forensic photography and such. I was looking at some theoretical issues with reference to magnification factors and digital technology in reference to a very theoretical/conceptual project I am working on. In this case, absolute accuracy isn't as important as is a defensible theoretical basis for calculating magnification--methodology--of the captured image, not the physical item photographed. </p>

<p>As I said, I was pretty amazed that my own thoughts, being a lifelong film shooter, went off into feeling that a rationale had to come out of the digital characteristics--and there are certain pulls still in that direction which truly do define digital as a different animal than is film.</p>

<p>Magnification is the ratio of linear size between any two images being compared.</p>

<p>Comparing the original object to the image on the sensor or film gives the magnification of the lens. Comparing the image on the sensor or film to the print gives the magnification of the enlargement. Comparing the original object to the print gives the total magnification.</p>

<p>Pixels have no more to do with it than film grain, with one exception that the magnification to the monitor at 100% zoom will be higher with more pixels. This does not affect the print.</p>

<p>As others have said, magnification is a ratio of linear dimensions. The ratio of surface area, while calculable, is irrelevant.</p>

<p>For example, if you have a sensor 20mm wide and make a print 200mm on that dimension, the magnification is 10x.</p>

<p>- Leigh</p>

 

<p>I think, bottom line, you will find that a digital images will enlarge better than a 35 mm negative. Somewhere around the 8MP sensors the DSLRs began surpassing film in "overall quality".</p>

<p>For example -- print a 36x24 inch print from a Canon 5D Mk II versus film at that size. Grain will be *much* more apparent in the 100 ISO film than the ISO 100 digital exposure at those large print sizes.</p>

<p>What's not said here are the dynamic ranges color negative film can capture over a modern CMOS.</p>

<p>Ken, that is certainly a good point. I have done comparisons from a 1dsmkIII @ 400 iso and an H2 with 400iso film where the digital capture is arguably better--even up sampled to match a 3200dpi film scan from the H2. But that isn't the issue here, quality, but enlargement factor.</p>

<p>As I said above, optical magnification--essentially sensor size--has been the standard in photography forever, however, I am still not totally convinced that the pixel density doesn't play a role in digital--as someone said, the file size. But that is all part of my project, to raise the issues.</p>

<blockquote>

<p>I am still not totally convinced that the pixel density doesn't play a role in digital--as someone said, the file size.</p>

</blockquote>

<p>The number of pixels may or may not significantly affect the <em>resolution </em>of the final print, depending mainly on the quality of the lens.</p>

<p>The number of pixels has absolutely no effect on the <em>magnification </em>of the print. More pixels do not cause the print to be bigger, therefore they do not magnify it. Magnification for a print from a 24mm x 36mm full frame sensor will usually range from 2.5x for a 2.5″ x 3.5″ print to 508x for a 60′ billboard.</p>

<p>I am trying to recreate your train of thought and I think I see where you are coming from. In the case of film you have two equivalent things to compare - a two dimensional image on a piece of film and a two dimensional image on a print. In the case of digital you don't have that, you have a file with some information content. The question then becomes what do you compare to the print - the information itself or some aspect of the thing that the information represents.</p>

<p>It seems to me that you can most defensibly use the term magnification in reference to an original equivalent state. In which case, I think the simplest thing to do is to compare the dimensions of the original sensor to the dimensions of the print and use optical magnification. The 10mp and 20mp files represent the same thing, albeit with different "densities" of information.</p>

<p>Thinking this way though it is pretty clear that there are various possible answers and you need to define fairly carefully what it is you are referring to as having been magnified in the print.</p>

<p>John A,<br>

At first I wasn't sure what you were getting at, but after re-reading your posts, I think I see. I think your initial thoughts on needing to know sensor size are correct if you are attempting to compute an approximate theoretical magnification factor. For example, assume we are photographing a 0.5 cm horizontal object with a macro lens at 2x (2:1). The lens will cast a 1 cm virtual image on the sensor plane. On a 5D Mark II the horizontal sensor dimension is 36 mm (3.6 cm) and records 5616 pixels or 5616 pix/3.6 cm = 1560 pix/cm recording your 1 cm virtual image. If you then printed this digital image at 360 dpi = 142 dpcm the object would print 1560 cm/142 dpcm = 11 cm or a magnification of 22X (11 cm/0.5 cm). Using this information you could reverse the math to calculate the theoretical size of a object in the picture (taking into account all the factors Walter Schroeder and others mention above.)</p>